What are the possible number of positive real, negative real, and complex zeros of . f(x) = 6x3 – 3x2 + 5x + 9? . . a)positive real: 1. negative real: 0. complex: 2. b)positive real: 2 or 0. negative real: 2 or 0. complex: 1. c)positive real: 2 or 0. negative real: 1. complex: 2 or 0. d)positive real: 1. negative real: 2 or 0. complex: 2 or 0.

C) The possible number of roots are

Positive Real root : 2 or 0.

Negative Real root: 1.

Complex root : 2 or 0

Step-by-step explanation:

Given :A cubic polynomial arranged in descending order

Therefore by  Descartes rules of signs . It has 2 sign change therefore it has 2 or zero positive real roots.

Now (change signs of coefficients of odd terms)

Now it has 1 sign change i.e. f(-x) has exactly 1 positive real roots. That means f(x) has exactly 1 negative real roots.

For complex roots , we know that it is a cubic polynomial i.e. it has exactly 3 roots and complex roots always occur in pair, Therefore it will have zero or 2 complex roots.

Therefore the possible number of roots are

Positive Real root : 2 or 0.

Negative Real root: 1.

Complex root : 2 or 0

Therefore option C is the correct answer .

answer: i'm not going to give you the answer, but some hints to work it out.

step-by-step explanation: so, if there is a straight line that means that the line angle is 180°. so we subtract the 2 angle measures provided, (15 and 38) from 180. so the answer is 180 - (38 + 15). (you have to solve the equation)

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation: